2.5 Absolute Value & Radicals

2.5 Absolute Value & Radicals

Absolute value and radical expressions frequently appear throughout algebra, calculus, and distance geometry. Understanding their algebraic properties, domain restrictions, simplification protocols, and rationalization techniques is vital for scoring high on the CLEP College Algebra exam.

---

Absolute Value: Definition & Properties

Geometrically, the absolute value of a real number $x$, written $|x|$, represents the magnitude or undirected distance between $x$ and $0$ on the real number line. Because distance cannot be negative, $|x| \ge 0$ for all real numbers $x \in \mathbb{R}$.

Piecewise Algebraic Definition

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

Core Algebraic Rule: If $x$ is negative, $-x$ is positive! For example, if $x = -7$, $|-7| = -(-7) = +7$.

Essential Algebraic Identities of Absolute Value

---

Radical Fundamentals & Principal Roots

For any integer index $n > 1$, the expression $\sqrt[n]{a}$ is a radical, where $n$ is the index and $a$ is the radicand.

• Even Roots ($n = 2, 4, 6, \dots$): $\sqrt[n]{a}$ exists in $\mathbb{R}$ only when $a \ge 0$. The non-negative output is called the principal $n$-th root.
• Odd Roots ($n = 3, 5, 7, \dots$): $\sqrt[n]{a}$ exists in $\mathbb{R}$ for all $a \in \mathbb{R}$. Odd roots preserve the algebraic sign of the radicand (e.g., $\sqrt[3]{-125} = -5$).

Core Operational Rules for Radicals

1. Product Rule: $\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$ (requires $a, b \ge 0$ if $n$ is even).
2. Quotient Rule: $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$ ($b \neq 0$).
3. Power of Radical: $(\sqrt[n]{a})^n = a$ (for $a \ge 0$ if $n$ even).
4. Radical of Power: $\sqrt[n]{a^n} = |a|$ if $n$ is even; $\sqrt[n]{a^n} = a$ if $n$ is odd.

---

Simplification of Radical Expressions

A radical expression is in simplified form if and only if:

1. The radicand has no factors with powers greater than or equal to the index $n$.
2. The radicand contains no fractions.
3. No radicals remain in the denominator of a fraction.

Worked Example: Multi-Variable Radical Simplification

Problem: Simplify the radical expression $\sqrt{252 x^9 y^6 z^{13}}$, assuming all variables represent positive real numbers ($x, y, z > 0$).

Step-by-Step Solution:

1. Factor numerical coefficient into prime factors / perfect squares: $252 = 36 \cdot 7 = 6^2 \cdot 7$

2. Decompose variable exponents into even powers and remainders:

   • $x^9 = x^8 \cdot x = (x^4)^2 \cdot x$
   • $y^6 = (y^3)^2$
   • $z^{13} = z^{12} \cdot z = (z^6)^2 \cdot z$

3. Group perfect squares under product rule: $\sqrt{(36 \cdot x^8 \cdot y^6 \cdot z^{12}) \cdot (7 \cdot x \cdot z)}$

4. Extract square roots of perfect powers: $\sqrt{36} \cdot \sqrt{x^8} \cdot \sqrt{y^6} \cdot \sqrt{z^{12}} \cdot \sqrt{7xz} = 6 x^4 y^3 z^6 \sqrt{7xz}$

---

Radical Combination & Multiplication

Radicals can only be combined via addition or subtraction if they are like radicals (having identical indices and identical radicands).

Worked Example: Radical Arithmetic & FOIL Expansion

Problem: Expand and fully simplify $(4\sqrt{3} - 3\sqrt{6})(2\sqrt{3} + 5\sqrt{6})$.

Step-by-Step Solution:

1. Apply FOIL Expansion:

   • First: $(4\sqrt{3})(2\sqrt{3}) = 8 \cdot 3 = 24$.
   • Outer: $(4\sqrt{3})(5\sqrt{6}) = 20\sqrt{18} = 20\sqrt{9 \cdot 2} = 20(3\sqrt{2}) = 60\sqrt{2}$.
   • Inner: $(-3\sqrt{6})(2\sqrt{3}) = -6\sqrt{18} = -6(3\sqrt{2}) = -18\sqrt{2}$.
   • Last: $(-3\sqrt{6})(5\sqrt{6}) = -15 \cdot 6 = -90$.

2. Combine integer constants and like radical terms: $(24 - 90) + (60\sqrt{2} - 18\sqrt{2}) = -66 + 42\sqrt{2}$

---

Rationalizing Fractional Denominators

1. Monomial Radical Denominators

Multiply numerator and denominator by a radical that creates a perfect $n$-th power in the denominator. $\frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$

2. Binomial Denominators (Conjugate Pairs)

The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$. Multiplying conjugate binomials eliminates radical terms via the Difference of Squares identity: $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - (\sqrt{b})^2 = a^2 - b$

Worked Example: Conjugate Rationalization

Problem: Rationalize the denominator of $\frac{6\sqrt{2}}{4 - \sqrt{10}}$.

Step-by-Step Solution:

1. Identify the conjugate of denominator $(4 - \sqrt{10})$: Conjugate is $(4 + \sqrt{10})$.
2. Multiply numerator and denominator by conjugate: $\frac{6\sqrt{2}(4 + \sqrt{10})}{(4 - \sqrt{10})(4 + \sqrt{10})}$
3. Simplify denominator: $(4)^2 - (\sqrt{10})^2 = 16 - 10 = 6$
4. Simplify numerator: $6\sqrt{2}(4) + 6\sqrt{2}(\sqrt{10}) = 24\sqrt{2} + 6\sqrt{20} = 24\sqrt{2} + 6(2\sqrt{5}) = 24\sqrt{2} + 12\sqrt{5}$
5. Divide by denominator $6$: $\frac{24\sqrt{2} + 12\sqrt{5}}{6} = 4\sqrt{2} + 2\sqrt{5}$